Let the equation of plane be ax + by + cz + d = 0 …. (1) Since the plane passes through the point A (0, 0, 0) and B(3, -1, 2), we have a × 0 + b × 0 + c × 0 + d = 0 ⇒ d = 0 …. (2) Similarly for point B (3, -1 , 2), a x 3 + b x (-1) + c x 2 + d = 0 3a – b + 2c = 0 (Using, d = 0) …(3) Given equation of the line is $\frac{x-4}{1}$ = $\frac{y+3}{-4}$ = $\frac{z-1}{7}$ We can also write the above equation as $\frac{x-4}{1}$ = $\frac{y-(-3)}{-4}$ = $\frac{x-(-1)}{7}$ The required plane is parallel to the above line. Therefore, a × 1 + b × (-4) + c × 7= 0 ⇒ a – 4b + 7c = 0 … (4) Cross multiplying equations (3) and (4), we obtain: $\frac{a}{(-1)\times 7-(-4)\times 2}$ = $\frac{b}{2\times 1-3\times 7}$ = $\frac{c}{3\times (-4)-1\times (-1)}$ ⇒ $\frac{a}{-7+8}$ = $\frac{b}{2-21}$ = $\frac{c}{-12+1}$ ⇒ $\frac{a}{1}$ = $\frac{b}{-19}$ = $\frac{c}{-11}$ = k ⇒ a = k , b = - 19k , c = - 11k Substituting the values of a, b and c in equation (1), we obtain the equation of plane as: kx - 19ky - 11kz + d = 0 ⇒ k (x - 19y - 11z) = 0 (From equation(2)) ⇒ x - 19y - 11z = 0 So, the equation of the required plane is x – 19y - 11z = 0